Solve this puzzle?

[p0ink]

You have two lines, AB and CD. AB || CD, and AB > CD

A line is drawn from A to C and from B to D. The two lines are extended until the intersect at a point E.

triangles ABE and CDE are similar

AB / BE = CD / DE

AB * DE = CD * BE

( AB - CD ) * AB * DE = ( AB - CD ) * CD * BE

AB * AB * DE - CD * AB * DE = AB * CD * BE - CD * CD * BE

AB * AB * DE - AB * CD * BE = CD * AB * DE - CD * CD * BE

AB * ( AB * DE - CD * BE ) = CD * ( AB * DE - CD * BE )

AB = CD - - - - how can this happen?

 

[Lestat]

you just blew my mind

 

[RADAR]

I'll get back after a few more drinks!

RADAR

 

[Lestat]

I'll get back after a few more drinks!

RADAR

bro I don't reccomend drinking with ulcers.

 

[RADAR]

bro I don't reccomend drinking with ulcers.

This is my first time tonight,besides it has healed a long while back, thanks for your concern tho!

RADAR

 

[Lestat]

This is my first time tonight,besides it has healed a long while back, thanks for your concern tho!

RADAR

just be careful man, alcohol, your age, and your medical history all add up to a bad accident waiting to happen. Don't do anything to worsen your chances.

 

[UA_Iron]

Only if BE is equal to DE

It's getting a little late. Point E could be at infinity - for all intensive purposes this would yield the same result.

 

[p0ink]

Only if BE is equal to DE

It's getting a little late. Point E could be at infinity - for all intensive purposes this would yield the same result.

point E is fixed, and it has an actual value rather than infinity.

 

[UA_Iron]

point E is fixed, and it has an actual value rather than infinity.

Interesting, I'll look at it again tomorrow.

 

[Mr. dB]

AB * DE = CD * BE

This line is false.

 

[UA_Iron]

This line is false.

I think that line is a product of line similitude.

 

[Mr. dB]

I think that line is a product of line similitude.

Ah, I misread it. I need new glasses. I was thinking it said AB * BE = CD * DE, which would be patently false. Never mind. Where are my reading glasses?

 

[blut wump]

You have two lines, AB and CD. AB || CD, and AB > CD

A line is drawn from A to C and from B to D. The two lines are extended until the intersect at a point E.

triangles ABE and CDE are similar

AB / BE = CD / DE

AB * DE = CD * BE

( AB - CD ) * AB * DE = ( AB - CD ) * CD * BE

AB * AB * DE - CD * AB * DE = AB * CD * BE - CD * CD * BE

AB * AB * DE - AB * CD * BE = CD * AB * DE - CD * CD * BE

AB * ( AB * DE - CD * BE ) = CD * ( AB * DE - CD * BE )

AB = CD - - - - how can this happen?

Since, AB * DE = CD * BE, we have that AB * DE - CD * BE = 0

All this has shown is that
AB * 0 = CD * 0, which has to be true.

Accordingly, the final cancellation is not valid. At the stage of multiplying though by (AB - CD) any two values (x - y) could have been chosen to 'show' that x = y.

 

[Bonkme2]

Since, AB * DE = CD * BE.

AB * DE - CD * BE = 0

Accordingly, the final cancellation is not valid. At the stage of multiplying though by (AB - CD) any two values (x - y) could have been chosen to 'show' that x = y.

He's right.
It's not so much that AB=CD in the last line...
It's AB*0 = CD*0 in the second to last line.

 

[blut wump]

You got in while I was editing

 

[blut wump]

To make it even clearer, if we put a value onto AB * DE, say 4. Of course, also CD * BE = 4.

We have AB * DE = CD * BE is equivalent to writing
4 = 4

Multiply through by (AB - CD)
(AB - CD) * 4 = (AB - CD) * 4
AB * 4 - CD * 4 = AB * 4 - CD * 4
AB * (4 - 4) = CD * (4 - 4)
AB = CD

 

[p0ink]

bing, bing. bing

we have a winner.